HDU 2224 The shortest path

The shortest path

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 2224
64-bit integer IO format: %I64d      Java class name: Main

 

There are n points on the plane, Pi(xi, yi)(1 <= i <= n), and xi < xj (i<j). You begin at P1 and visit all points then back to P1. But there is a constraint: 
Before you reach the rightmost point Pn, you can only visit the points those have the bigger x-coordinate value. For example, you are at Pi now, then you can only visit Pj(j > i). When you reach Pn, the rule is changed, from now on you can only visit the points those have the smaller x-coordinate value than the point you are in now, for example, you are at Pi now, then you can only visit Pj(j < i). And in the end you back to P1 and the tour is over.
You should visit all points in this tour and you can visit every point only once.

 

Input

The input consists of multiple test cases. Each case begins with a line containing a positive integer n(2 <= n <= 200), means the number of points. Then following n lines each containing two positive integers Pi(xi, yi), indicating the coordinate of the i-th point in the plane.

 

Output

For each test case, output one line containing the shortest path to visit all the points with the rule mentioned above.The answer should accurate up to 2 decimal places.

 

Sample Input

3
1 1
2 3
3 1

Sample Output

6.47

Hint: The way 1 - 3 - 2 - 1 makes the shortest path.

解题：双调TSP

#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 500;
typedef long long LL;
double dp[maxn][maxn],ds[maxn][maxn];
struct Point {
    int x,y;
    bool operator<(const Point &t) const {
        return x < t.x;
    }
} p[maxn];
double calc(int a,int b) {
    LL x = p[a].x - p[b].x;
    LL y = p[a].y - p[b].y;
    double tmp = x*x + y*y;
    return sqrt(tmp);
}

int main() {
    int n;
    while(~scanf("%d",&n)) {
        for(int i = 1; i <= n; ++i)
            scanf("%d%d",&p[i].x,&p[i].y);
        sort(p,p+n);
        for(int i = 1; i <= n; ++i)
            for(int j = 1; j <= n; ++j)
                ds[i][j] = calc(i,j);
        dp[2][1] = ds[1][2];
        for(int i = 3; i <= n; ++i) {
            dp[i][i-1] = INF;
            for(int j = 1; j < i - 1; j++) {
                dp[i][j] = dp[i-1][j] + ds[i][i-1];
                dp[i][i-1] = min(dp[i][i-1],dp[i-1][j] + ds[j][i]);
            }
        }
        printf("%.2f
",dp[n][n-1]+ds[n-1][n]);
    }
    return 0;
}